Question

guys plz do answer me for the 3rd question.... irrelevant answers will be reported on the spot....​

Answer 1

Answer:

y² - 4y + 3 = 0

y² - 3y - y + 3 = 0

y ( y - 3) - 1 ( y - 3 )= 0

( y - 1 ) = 0 , ( y - 3 ) = 0

so,

y = 1 or Y = 3 ......

Thanks

mark brainliest tyvm!!!

Answer 2

Method 1 ⇒ Factorization

Let's solve your equation step-by-step.

$y^{2}-4y+3=0$

Step 1: Factor left side of equation.

⇒ $(y-1)(y-3)=0$

Step 2: Set factors equal to 0.

⇒ $y-1=0 \ or \ y-3=0$

⇒ $y= 3 \ or \ y=1$

$\rule{300}{1}$

Method 2 ⇒ With Quadratic Formula

Let's solve your equation step-by-step.

$y^{2}-4y+3=0$

For this equation: a = 1, b = -4, c = 3

⇒ $1y^{2}+-4y+3=0$

Step 1: Use quadratic formula with a = 1, b = -4, c = 3.

⇒ $y=\dfrac{-b\±\sqrt{b^{2}-4ac}}{2a}$

⇒ $y=\dfrac{-(-4)\±\sqrt{(-4)^{2}-4(1)(3)}}{2(1)}$

⇒ $y=\dfrac{4\±\sqrt{16-4(3)}}{2(1)}$

⇒ $y=\dfrac{4\±\sqrt{16-12}}{2(1)}$

⇒ $y=\dfrac{4\±\sqrt{4}}{2}$

⇒ $y= 3 \ or \ y=1$

$\rule{300}{1}$

$\bf \therefore y=3\ or \ y = 1 \ in \ the\ equation =>\ y^{2}-4y+3=0$

$\rule{300}{1}$